Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T& num) {
  bool start = false, neg = false;
  char c;
  num = 0;
  while ((c = getchar()) != EOF) {
    if (c == '-')
      start = neg = true;
    else if (c >= '0' && c <= '9') {
      start = true;
      num = num * 10 + c - '0';
    } else if (start)
      break;
  }
  if (neg) num = -num;
}
const int maxn = 105;
const int MAX = (int)(1e6) + 5;
int ai[maxn][maxn];
int R[maxn], C[maxn];
int n, m;
int tot;
int res[MAX], opt[MAX];
int main() {
  read(n);
  read(m);
  int su = 0;
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (m); j++) {
      read(ai[i][j]);
      su += ai[i][j];
      if (!ai[i][j]) R[i] = C[j] = 1;
    }
  while (1) {
    if (n >= m) {
      int k = 0;
      for (int j = (1); j <= (m); j++)
        if (!C[j]) k = j;
      if (k) {
        ++tot;
        opt[tot] = 1;
        res[tot] = k;
        for (int i = (1); i <= (n); i++) {
          if ((--ai[i][k]) == 0) R[i] = C[k] = 1;
        }
        su -= n;
      } else {
        for (int i = (1); i <= (n); i++)
          if (!R[i]) k = i;
        if (k) {
          ++tot;
          opt[tot] = 2;
          res[tot] = k;
          for (int j = (1); j <= (m); j++) {
            if ((--ai[k][j]) == 0) R[k] = C[j] = 1;
          }
          su -= m;
        } else {
          if (su) return printf("-1\n"), 0;
          break;
        }
      }
    } else {
      int k = 0;
      for (int i = (1); i <= (n); i++)
        if (!R[i]) k = i;
      if (k) {
        ++tot;
        opt[tot] = 2;
        res[tot] = k;
        for (int j = (1); j <= (m); j++) {
          if ((--ai[k][j]) == 0) R[k] = C[j] = 1;
        }
        su -= m;
      } else {
        for (int j = (1); j <= (m); j++)
          if (!C[j]) k = j;
        if (k) {
          ++tot;
          opt[tot] = 1;
          res[tot] = k;
          for (int i = (1); i <= (n); i++) {
            if ((--ai[i][k]) == 0) R[i] = C[k] = 1;
          }
          su -= n;
        } else {
          if (su) return printf("-1\n"), 0;
          break;
        }
      }
    }
  }
  printf("%d\n", tot);
  for (int i = (1); i <= (tot); i++) {
    if (opt[i] == 2)
      printf("row ");
    else
      printf("col ");
    printf("%d\n", res[i]);
  }
  return 0;
}